1. How many cubes of 10 cm edge can be put in a cubic box of 1 m edge ?
(1) 10 (2) 100
(3) 1000 (4) 10000
Ans. The volume of small cube of 10 cm is = [tex]{10^3} = 1000[/tex] [tex]c{m^3}[/tex]
The volume of 1 m = 100 cm cubic box is = [tex]{100^3} = 1000000[/tex] [tex]c{m^3}[/tex]
Number of cubes are [tex]\frac{{1000000}}{{1000}} = 1000[/tex]
2. Next terms of the series 198, 194, 185, 169 .....
(1) 92 (2)112
(3)136 (4)144
Ans. The different between 1st and 2nd term is ( 198 - 194 ) = 4=[tex]{2^2}[/tex]
The different between 2nd and 3rd term is ( 194 - 185 ) = 9 = [tex]{3^2}[/tex]
The different between 3rd and 4th term is ( 185 - 169 ) = 16 = [tex]{4^2}[/tex]
Then we can say the different between 4th and 5th term is [tex]{5^2} = 25[/tex]
Then the next term is 169 - 25 = 144
3. Which fraction comes next in the sequence
[tex]\frac{1}{2},\frac{3}{4},\frac{5}{8},\frac{7}{{16}}[/tex]
(1) [tex]\frac{9}{{32}}[/tex] (2) [tex]\frac{{10}}{{17}}[/tex]
(3) [tex]\frac{{11}}{{34}}[/tex] (4) [tex]\frac{{12}}{{35}}[/tex]
Ans. The sequence is [tex]\frac{1}{2},\frac{3}{4},\frac{5}{8},\frac{7}{{16}}[/tex]
Then
[tex]\begin{array}{l}
\frac{1}{2},\frac{3}{4},\frac{5}{8},\frac{7}{{16}}\\
\frac{1}{{{2^1}}},\frac{{1 + 2}}{{{2^2}}},\frac{{1 + 2 + 2}}{{{2^3}}},\frac{{1 + 2 + 2 + 2}}{{{2^4}}}
\end{array}[/tex]
Now the next term will be [tex]\frac{{1 + 2 + 2 + 2 + 2}}{{{2^5}}} = \frac{9}{{32}}[/tex]
4. The last day of century cannot be
(1) Monday (2) Wednesday
(3) Tuesday (4) Friday
Ans. 100 years contain 5 odd days.So last day of 1st century is Friday.
200 years contain [tex](5 \times 2) \equiv 3[/tex] odd days. So last day of 2nd century is Wednesday .
300 years contain [tex](5 \times 3) = 15 \equiv 1[/tex] odd day. So last day of 3rd century is Monday.
400 years contain 0 odd day. So the last day of 4th century is Sunday.
this cycle is repeated. So last day of century cannot be Tuesday, Thursday or Saturday.
5. The area of a square is equal to the area of a circle. The ratio between the side of a square and the radius of the circle is
(1) [tex]\sqrt \pi :1[/tex] (2) [tex]1:\sqrt \pi [/tex]
(3) [tex]1:\pi [/tex] (4) [tex]\pi :1[/tex]
Ans. Let the one side of the square is a. Then the area of the square is [tex]{a^2}[/tex].
Let the radius of the circle is r. Then the area of the circle is [tex]\pi {r^2}[/tex].
Now we can write
[tex]\begin{array}{l}
{a^2} = \pi {r^2}\\
\Rightarrow a = \sqrt \pi r\\
\Rightarrow \frac{a}{r} = \sqrt \pi \\
\Rightarrow a:r = \sqrt \pi :1
\end{array}[/tex]
6. Two trains, each 100 m long moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast as the other, then the speed of the faster train is
(1) 40 km/hr (2) 50 km/hr
(3) 60 km/hr (4) 70 km/hr
Ans. The speed of the slower train is u m/sec, then the speed of the faster train is 2u m/sec.
The trains moving in opposite direction,then the relative velocity is (u + 2u ) m/sec = 3u m/sec
Then ( 100 + 100 ) m = 200 m cross time of each train is [tex]\frac{{200}}{{3u}}[/tex] sec.
Now we can write [tex]8 = \frac{{200}}{{3u}} \Rightarrow u = \frac{{200}}{{3 \times 8}} \Rightarrow u = \frac{{25}}{3}[/tex] m/sec
Then the speed of the faster train is [tex]2u = 2 \times \frac{{25}}{3}[/tex] m/sec or [tex]2 \times \frac{{25}}{3} \times \frac{{3600}}{{1000}} = 60[/tex] km/hr.
7. In a river a man takes 3 hour in rowing 3 km up-stream or 15 km down-stream, the speed of the current is
(1) 2 km/hr (2) 4 km/hr
(3) 6 km/hr (4) 9 km/hr
Ans. Let the speed of the current is u km/hr and the speed of the man is v km/hr in no stream.
Then the man rowing ( v - u ) km in 1 hour in up-stream and ( v + u ) km in 1 hour in down-stream.
Then we can write 3( v - u ) = 3 and 3( v + u ) =15
[tex]\begin{array}{l}
3(v - u) = 3 \Rightarrow v - u = 1............(1)\\
3(v + u) = 15 \Rightarrow v + u = 5...........(2)\\
(2) - (1)\\
2u = 4 \Rightarrow u = 2
\end{array}[/tex]
The speed of the current is 2 km/hr.
8. In what ratio the water be mixed with milk to gain [tex]16\frac{2}{3}\% [/tex] on selling the mixture at cost price?
(1) 1 : 6 (2) 2:3
(3) 4 : 3 (4) 6 : 1
Ans. Let the quantity of the milk is x liters and the amount of the water is y liters.Then C.P of x liters =S.P of ( x + y ) liters.
Gain = [tex]16\frac{2}{3}\% = \frac{{50}}{3}\% [/tex]
[tex]\begin{array}{l}
S.P = C.P(1 + \frac{{Gain\% }}{{100}})\\
\Rightarrow x + y = x(1 + \frac{{50}}{{300}})\\
\Rightarrow x + y = x\frac{{35}}{{30}}\\
\Rightarrow 30x + 30y = 35x\\
\Rightarrow 30y = 5x\\
\Rightarrow \frac{y}{x} = \frac{1}{6}\\
\Rightarrow y:x = 1:6
\end{array}[/tex]
9. Simple interest of Rs. 16,250 at 8% per annum for 73 days is
(1) Rs. 460 (2) Rs. 260
(3) Rs. 560 (3) Rs.660
Ans. We know [tex]S.I = \frac{{P \times R \times T}}{{100}}[/tex]
Here P = Rs.16,250 R = 8% T = [tex]\frac{{73}}{{365}}[/tex] years
Then [tex]S.I = \frac{{16250 \times 8 \times \frac{{73}}{{365}}}}{{100}} = 260[/tex]
S.I = Rs. 260
10. The diagonals of two squares are in the ratio 5 : 2 of their area is
(1) 5 : 2 (2) 25 : 4
(3) 125 : 8 (4) 4 : 25
Ans. Let the sides of two squares are x and y
Then the diagonals of them are [tex](\sqrt {{x^2} + {x^2}} = \sqrt 2 x),(\sqrt {{y^2} + {y^2}} = \sqrt 2 y)[/tex]
Then we can write [tex]\frac{{\sqrt 2 x}}{{\sqrt 2 y}} = \frac{5}{2} \Rightarrow \frac{x}{y} = \frac{5}{2} \Rightarrow x = \frac{5}{2}y[/tex]
The area of them are [tex]{x^2}[/tex] and [tex]{y^2}[/tex]
Then [tex]\frac{{{x^2}}}{{{y^2}}} = \frac{{{{(\frac{5}{2})}^2}{y^2}}}{{{y^2}}} = \frac{{25}}{4}[/tex]
The ratio of the areas is 25 : 4
11. Each side of a rhombus is 5 cm. Its area is
(1) 25 [tex]c{m^2}[/tex] (2) 23 [tex]c{m^2}[/tex]
(3) 24 [tex]c{m^2}[/tex] (4) data inadequate
Ans. We know the area of rhombus is [tex]\frac{1}{2} \times {d_1} \times {d_2}[/tex] , where [tex]{d_1},{d_2}[/tex] are two diagonals of rhombus.
Here is given each side of rhombus is 5 cm.
Then the data is inadequate to find.
12. A man bought 5 shirt at Rs. 450 each, 4 trousers at Rs. 750 each and 12 pairs of shoes at Rs. 750 each. The average expenditure per article is
(1) Rs 678.50 (2) Rs 800
(3) Rs 900 (4) Rs 1000
Ans:
[tex]\begin{array}{l}
\frac{{(5 \times 450) + (4 \times 750) + (12 \times 750)}}{{5 + 4 + 12}}\\
= \frac{{2250 + 3000 + 9000}}{{21}}\\
= \frac{{14250}}{{21}}\\
= 678.57
\end{array}[/tex]
13. 45% of 280 + 28% of 450 = ?
(1) 152 (2) 252
(3) 354 (4) 454
Ans.
[tex]\begin{array}{l}
\frac{{45}}{{100}} \times 280 + \frac{{28}}{{100}} \times 450\\
= \frac{{12600}}{{100}} + \frac{{12600}}{{100}}\\
= 126 + 126\\
= 252
\end{array}[/tex]
14. The radius of a circle is increased by 1%. Then percentage increased in area is
(1) 1% (2) 1.01%
(3) 2% (4) 2.01%
Ans. Let the radius of the circle is r unit. Then area of it is [tex]\pi {r^2}uni{t^2}[/tex].
The radius of a circle is increased by 1%. Then new radius is [tex]\frac{{101r}}{{100}}[/tex] unit and area of it is [tex]\pi {(\frac{{101r}}{{100}})^2} = \pi \frac{{10201}}{{10000}}{r^2}uni{t^2}[/tex]
The increased in area is [tex]\pi \frac{{10201}}{{10000}}{r^2} - \pi {r^2} = \pi (\frac{{10201 - 10000}}{{10000}}){r^2} = \pi \frac{{201}}{{10000}}{r^2}uni{t^2}[/tex]
Then percentage increased in area is [tex]\frac{{201 \times 100}}{{10000}} = 2.01[/tex]
15. A dishonest dealer claim to sell his good at the cost price but use a false weight 900 gm for 1 kg. His gain percentage is
(1) 13% (2) [tex]11\frac{1}{9}\% [/tex]
(3) 11.25% (4) [tex]12\frac{1}{9}\% [/tex]
Ans. The dishonest dealer weight gain in 900 gm is ( 1000 - 900 ) gm = 100 gm.
The gain percentage in 900 gm is [tex]\frac{{100}}{{900}} \times 100 = \frac{{100}}{9} = 11\frac{1}{9}[/tex]
His gain percentage is [tex]11\frac{1}{9}\% [/tex]
16. By selling a table for Rs. 350 instead of Rs. 400 loss percent increased by 5%; The cost price of table is
(1) Rs. 435 (2) Rs. 417.50
(3) Rs. 1,000 (4) Rs. 1,050
Ans. Let the cost price of the table is Rs. x .
We know the loss %
[tex]\begin{array}{l}
\left( {\frac{{C.P. - S.P.}}{{C.P.}}} \right) \times 100\% \\
\Rightarrow \left( {\frac{{x - 350}}{x}} \right) - \left( {\frac{{x - 400}}{x}} \right) = \frac{5}{{100}}\\
\Rightarrow \frac{{x - 350 - x + 400}}{x} = \frac{5}{{100}}\\
\Rightarrow 5x = 5000\\
\Rightarrow x = 1000
\end{array}[/tex]
Then the cost of the table is Rs.1000
17. A constable is 114 m behind a thief. The constable runs 21 m and thief 15 m in a minute. In what time will the constable catch the thief?
(1) 16 minutes (2) 17 minutes
(3) 18 minutes (4) 19 minutes
Ans. The constable covers 21 m in 1 minutes and thief covers 15 m in 1 minutes. Then constable covers ( 21 - 15 ) m = 6 m more than thief in 1 minutes.
Now constable covers 114 m in [tex]\frac{{114}}{6} = 19[/tex] minutes.
In 19 minutes will the constable catch the thief.
18. [tex]{\left[ {{{\left\{ {{{\left( { - \frac{1}{2}} \right)}^2}} \right\}}^{ - 2}}} \right]^{ - 1}}[/tex]
(1) 16 (2) [tex]\frac{1}{{16}}[/tex]
(3) 4 (4) [tex]\frac{1}{4}[/tex]
Ans.
[tex]\begin{array}{l}
{\left[ {{{\left\{ {{{\left( { - \frac{1}{2}} \right)}^2}} \right\}}^{ - 2}}} \right]^{ - 1}}\\
= {\left[ {{{\left\{ {\frac{1}{4}} \right\}}^{ - 2}}} \right]^{ - 1}}\\
= {\left[ {16} \right]^{ - 1}}\\
= \frac{1}{{16}}(ans)
\end{array}[/tex]
19. The number of the digits of the square root of 0.00059049 is
(1) 5 (2) 6
(3) 4 (4) 3
Ans. The square root of 0.00059049 is
[tex]\begin{array}{l}
\sqrt {0.00059049} \\
= \sqrt {\frac{{59049}}{{100000000}}} \\
= \frac{{\sqrt {59049} }}{{10000}}\\
= \frac{{243}}{{10000}}\\
= 0.0243
\end{array}[/tex]
Total number of digits is 4
20. 3 men or 5 women can do a work in 12 days. How long will 6 men and 5 women take to finish the work?
(1) 4 days (2) 10 days
(3) 15 days (4) 20 days
Ans. From the question we can write 3 men = 5 women. So 6 men and 5 women are total ( 6 + 3 ) men or 9 men.
Then 3 men can do a work in 12 days
Now 1 men can do a work in [tex]12 \times 3 = 36[/tex] days
Therefor 9 men can do this work in [tex]36 \div 9 = 4[/tex] days
21. Two pipes A and B can fill a cistern in 6 minutes and 7 minutes respectively. Both the pipes are opened alternately for 1 minute each what time will they fill the cistern ?
(1) 5 minutes (2) [tex]5\frac{2}{3}[/tex] minutes
(3) [tex]6\frac{3}{7}[/tex] minutes (4) [tex]1\frac{1}{4}[/tex] minutes
Ans. Let the volume of the cistern is x.
Then A pipes fill the water in 1 minutes is [tex]\frac{x}{6}[/tex] volume and B pipes fill the water in 1 minutes is [tex]\frac{x}{7}[/tex] volume of cistern.
The volume of cistern fill the water in 2 minutes is [tex]\left( {\frac{x}{6} + \frac{x}{7}} \right) = \left( {\frac{{7 + 6}}{{42}}} \right)x = \frac{{13}}{{42}}x[/tex]
Therefor [tex]\frac{{13}}{{42}}x[/tex] volume of cistern fill of water in 2 minutes.
The part of cistern fill A and B in [tex]2 \times 3 = 6[/tex] minutes = [tex]3 \times \frac{{13x}}{{42}} = \frac{{39x}}{{42}}[/tex] volume.
Part of cistern left =[tex]\left( {x - \frac{{39x}}{{42}}} \right)volume = \frac{{3x}}{{42}}volume = \frac{x}{{14}}volume[/tex]
Now it turn of A
[tex]\frac{x}{6}[/tex] part fill by A in 1 minute.
1 part fill by A in [tex]\frac{6}{x}[/tex] minute.
Then [tex]\frac{x}{{14}}[/tex] part fill by A in [tex]\frac{6}{x} \times \frac{x}{{14}} = \frac{6}{{14}}[/tex] minutes
Total time taken
[tex]\left( {6 + \frac{6}{{14}}} \right) = 6 \times \frac{{15}}{{14}} = \frac{{45}}{7} = 6\frac{3}{7}[/tex]
22. How many times in a day, the hands of a clock are straight ?
(1) 22 (2) 24
(3) 44 (4) 48
Ans. Hands of a clock point in opposite directions i,e in the same straight lines in 11 times in every 12 hours. Hence in a day hands of clock are 22 times in straight lines.
23. The reflex angle between the hands of a clock at 10.25 is
(1) [tex]{180^0}[/tex] (2) [tex]192\frac{{{1^0}}}{2}[/tex]
(3) [tex]{195^0}[/tex] (4) [tex]197\frac{{{1^0}}}{2}[/tex]
Ans. Angle subtended by hour hand in [tex]10\frac{{25}}{{60}} = \frac{{125}}{{12}}[/tex] hrs = [tex]\left( {\frac{{{{360}^0}}}{{12}} \times \frac{{125}}{{12}}} \right) = 312\frac{{{1^0}}}{2}[/tex]
Angle subtended by minute hand in 25 min = [tex]\left( {\frac{{{{360}^0}}}{{60}} \times 25} \right) = {150^0}[/tex]
Reflex angle
[tex]\begin{array}{l}
{360^0} - {\left( {312.5 - 150} \right)^0}\\
= {360^0} - {162.5^0}\\
= {197.5^0}\\
= 197\frac{{{1^0}}}{2}
\end{array}[/tex]
24. The sum 5 + 6 + 7 + 8 + ............ + 19 = ?
(1) 150 (2) 170
(3) 180 (4) 190
Ans. We know the sum of A.P is
[tex]\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right][/tex]
Where a = 5, n = 15, d = 1
Sum [tex]\frac{{15}}{2}\left[ {2 \times 5 + \left( {15 - 1} \right) \times 1} \right] = \frac{{15}}{2}\left( {10 + 14} \right) = \frac{{15 \times 24}}{2} = 180[/tex]
25. [tex]0.3 + 0.4 + 0.7 + 0.8 = ?[/tex]
(1) 2.4444 (2)2.44
(3) 2.444 (4) 2.4
Ans.
[tex]\begin{array}{l}
{0.3^0} + {0.4^0} + {0.7^0} + {0.8^0}\\
= \frac{3}{9} + \frac{4}{9} + \frac{7}{9} + \frac{8}{9}\\
= \frac{{3 + 4 + 7 + 8}}{9}\\
= \frac{{22}}{9}\\
= 2.444\\
= {2.4^0}
\end{array}[/tex]
26. If the ratio of three numbers is 3 : 4 : 5 and their LCM is 1200, then the smaller number is
(1) 60 (2) 80
(3) 100 (4) 120
Ans. Let the three numbers are 3x , 4x and 5x.
Therefor the LCM of them is [tex]3 \times 4 \times 5 \times x = 60x[/tex]
Then we can say
[tex]\begin{array}{l}
60x = 1200\\
\Rightarrow x = 20
\end{array}[/tex]
Then smaller number is [tex]3 \times 20 = 60[/tex]
27. The product of two numbers is 1575 and its division is [tex]\frac{9}{7}[/tex]. Then the two numbers are
(1) 45 , 35 (2) 81 , 63
(3) 35 , 27 (4) 36 , 35
Ans. Let two numbers are x and y.
Then [tex]x \times y = 1575.........................(i)[/tex]
and [tex]\frac{x}{y} = \frac{9}{7}.....................................(ii)[/tex]
From (ii) we get [tex]\frac{x}{y} = \frac{9}{7} \Rightarrow x = \frac{{9y}}{7}................................\left( {iii} \right)[/tex]
put (iii) in (i)
[tex]\begin{array}{l}
\frac{{9y}}{7} \times y = 1575\\
\Rightarrow {y^2} = \frac{{1575 \times 7}}{9} = 1225\\
\Rightarrow y = \sqrt {1225} = \pm 35
\end{array}[/tex]
Then another number will be [tex]1575 \div \left( { \pm 35} \right) = \pm 45[/tex]
28. [tex]\sqrt {2025} + \sqrt {441} + \sqrt {169} = ?[/tex]
(1) 59 (2) 69
(3) 79 (4) 89
Ans.
[tex]\begin{array}{l}
\sqrt {2025} + \sqrt {441} + \sqrt {169} \\
= \sqrt {5 \times 5 \times 9 \times 9} + \sqrt {9 \times 7 \times 7} + \sqrt {13 \times 13} \\
= 5 \times 9 + 3 \times 7 + 13\\
= 45 + 21 + 13\\
= 79
\end{array}[/tex]
29. The ratio of two numbers is 5 : 8 and their difference is 69. Then the two numbers are
(1) 69 , 128 (2) 115 , 184
(3) 43 , 112 (4) 128 , 197
Ans. Let the two numbers are 5x and 8x. Then we can write
8x - 5x = 69
or, 3x = 69
or, x = 23
Therefor two numbers are [tex]\left( {5 \times 23} \right) = 115[/tex] and [tex]\left( {8 \times 23} \right) = 184[/tex]
30. If [tex]A:B = \frac{1}{6}:\frac{1}{5},B:C = \frac{1}{4}:\frac{1}{3},C:D = \frac{1}{3}:\frac{1}{5}[/tex] then A : D is
(1) [tex]\frac{1}{{24}}:\frac{1}{{25}}[/tex] (2) [tex]\frac{1}{{27}}:\frac{1}{{25}}[/tex]
(3) [tex]\frac{1}{{24}}:\frac{1}{{29}}[/tex] (4) None of the above
Ans. [tex]\frac{A}{B} = \frac{5}{6},\frac{B}{C} = \frac{3}{4},\frac{C}{D} = \frac{5}{3}[/tex]
Then
[tex]\begin{array}{l}
\frac{A}{B} \times \frac{B}{C} \times \frac{C}{D} = \frac{5}{6} \times \frac{3}{4} \times \frac{5}{3}\\
\Rightarrow \frac{A}{D} = \frac{{25}}{{24}}\\
\Rightarrow A:D = \frac{1}{{24}}:\frac{1}{{25}}
\end{array}[/tex]
31. [tex]\frac{{{{\left( {6.5} \right)}^2} - {{\left( {3.15} \right)}^2}}}{{\left( {6.5 + 3.15} \right)}} = ?[/tex]
(1) 3.5 (2) 3.51
(3) 3.52 (4) 3.35
Ans.
[tex]\begin{array}{l}
\frac{{{{\left( {6.5} \right)}^2} - {{\left( {3.15} \right)}^2}}}{{\left( {6.5 + 3.15} \right)}}\\
= \frac{{\left( {6.5 + 3.15} \right) \times \left( {6.5 - 3.15} \right)}}{{\left( {6.5 + 3.15} \right)}}\\
= \left( {6.5 - 3.15} \right)\\
= 3.35
\end{array}[/tex]u
32. The compound interest on Rs. 2000 for 2 years at 8%per annum is
(1) Rs. 220.80 (2) Rs. 232.80
(3) Rs. 332.80 (4) Rs. 532.80
Ans. We know that [tex]C.I. = P\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^n} - 1} \right][/tex]
Where P = 2000 , r = 8 , n = 2
[tex]\begin{array}{l}
C.I. = 2000\left[ {{{\left( {1 + \frac{8}{{100}}} \right)}^2} - 1} \right]\\
= 2000\left[ {{{\left( {\frac{{108}}{{100}}} \right)}^2} - 1} \right]\\
= 2000\left( {\frac{{108}}{{100}} + 1} \right)\left( {\frac{{108}}{{100}} - 1} \right)\\
= 2000 \times \frac{{208}}{{100}} \times \frac{8}{{100}}\\
= 332.80
\end{array}[/tex]
33. The rational numbers lying between [tex]\frac{1}{4}[/tex] and [tex]\frac{3}{4}[/tex] are
(1) [tex]\frac{9}{{40}},\frac{{31}}{{41}}[/tex] (2) [tex]\frac{{13}}{{50}},\frac{{264}}{{350}}[/tex]
(3) [tex]\frac{{63}}{{250}},\frac{{187}}{{250}}[/tex] (4) [tex]\frac{{262}}{{1000}},\frac{{752}}{{1000}}[/tex]
Ans. [tex]\frac{1}{4} \times \frac{{250}}{{250}} = \frac{{250}}{{1000}},\frac{3}{4} \times \frac{{250}}{{250}} = \frac{{750}}{{1000}}[/tex]
Now from the option
[tex]\begin{array}{l}
\frac{{63}}{{250}} \times \frac{4}{4} = \frac{{252}}{{1000}},\frac{{187}}{{250}} \times \frac{4}{4} = \frac{{748}}{{1000}}\\
i.e\frac{1}{4} < \frac{{63}}{{250}} < \frac{3}{4},\frac{1}{4} < \frac{{187}}{{250}} < \frac{3}{4}
\end{array}[/tex]
Then option (3) lying between [tex]\frac{1}{4}[/tex] and [tex]\frac{3}{4}[/tex]
34. [tex]{\left( {77 + \frac{1}{{77}}} \right)^2} - {\left( {77 - \frac{1}{{77}}} \right)^2} = ?[/tex]
(1) 2 (2) 4
(3) 1 (4) 77
Ans.
[tex]\begin{array}{l}
{\left( {77 + \frac{1}{{77}}} \right)^2} - {\left( {77 - \frac{1}{{77}}} \right)^2}\\
= 4 \times 77 \times \frac{1}{{77}}\left[ {4ab = {{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}} \right]\\
= 4
\end{array}[/tex]
35. By what least number must 21600 be multiplied to make it a perfect cube?
(1) 6 (2) 10
(3) 30 (4) 60
Ans. [tex]21600 = 3 \times 3 \times 3 \times 2 \times 2 \times 2 \times 10 \times 10[/tex]
At least 10 multiplied to this number to make a perfect cube.
36. If P = 50% of Q and Q = 50% of R, then P : Q : R = ?
(1) 1 : 2 : 4 (2) 1 : 4 : 2
(3) 4 : 2 : 1 (4) 2 : 1 : 4
Ans. P = 50% of Q, then [tex]P = \frac{{50}}{{100}}Q = \frac{1}{2}Q[/tex] ...........................(i)
and Q = 50% of R, then [tex]Q = \frac{{50}}{{100}}R = \frac{1}{2}R[/tex] .............................(ii)
From (i) 2P=Q , and (ii) 2Q=R
Then [tex]2 \times 2P = R \Rightarrow 4P = R[/tex]
Therefor P : Q : R = P : 2P : 4P =1 : 2 ; 4
37. The average of first 100 natural numbers is
(1) 50 (2) 50.5
(3) 51 (4) 51.5
Ans. We know that the average of first n natural numbers is [tex]\frac{{\left( {n + 1} \right)}}{2}[/tex]
Then the required average is [tex]\frac{{\left( {100 + 1} \right)}}{2} = \frac{{101}}{2} = 50.5[/tex]
38. If the sum of two numbers is 10 and the sum their reciprocals is [tex]\frac{5}{{12}}[/tex] the numbers would be
(1) ( 8 , 2 ) (2) ( 6 , 4 )
(3) ( 7 , 3 ) (4) ( 9 , 1 )
Ans. Let the two numbers are x and y .
Then x + y = 10 ......................(i) and [tex]\frac{1}{x} + \frac{1}{y} = \frac{5}{{12}}..........................(ii)[/tex]
Now from (i)
x = 10 - y...............................(iii)
use (iii) in (ii) we get
[tex]\begin{array}{l}
\frac{1}{{10 - y}} + \frac{1}{y} = \frac{5}{{12}}\\
\Rightarrow \frac{{y + 10 - y}}{{y\left( {10 - y} \right)}} = \frac{5}{{12}}\\
\Rightarrow \frac{{10}}{{y\left( {10 - y} \right)}} = \frac{5}{{12}}\\
\Rightarrow 24 = 10y - {y^2}\\
\Rightarrow {y^2} - 10y + 24 = 0\\
\Rightarrow {y^2} - \left( {6 + 4} \right)y + 24 = 0\\
\Rightarrow y\left( {y - 6} \right) - 4\left( {y - 6} \right) = 0\\
\Rightarrow \left( {y - 6} \right)\left( {y - 4} \right) = 0\\
\Rightarrow y = 6,4
\end{array}[/tex]
If y = 6 , then x = 4 and if y = 4 , then x = 6
39. The product of two successive numbers is 1980. The smaller number is
(1) 34 (2) 35
(3) 44 (4) 45
Ans. Let n and (n+1) are two successive numbers. Then
[tex]\begin{array}{l}
n\left( {n + 1} \right) = 1980\\
\Rightarrow {n^2} + n - 1980 = 0\\
\Rightarrow {n^2} + \left( {45 - 44} \right)n - 1980 = 0\\
\Rightarrow {n^2} + 45n - 44n - 1980 = 0\\
\Rightarrow n\left( {n + 45} \right) - 44\left( {n + 45} \right) = 0\\
\Rightarrow \left( {n - 44} \right)\left( {n + 45} \right) = 0\\
\Rightarrow n = 44,\left( { - 45} \right)
\end{array}[/tex]
Then smaller number is 44
40. [tex]\frac{4}{5}[/tex] of certain number is 64. Half of the number
(1) 32 (2) 40
(3) 80 (4) 16
Ans. Let the certain number is x. Then
[tex]\begin{array}{l}
\frac{4}{5}x = 64\\
\Rightarrow x = 64 \times \frac{5}{4} = 80
\end{array}[/tex]
Therefor the half of x is [tex]80 \div 2 = 40[/tex]
41. Which is greater [tex]\sqrt 2 [/tex] or [tex]3\sqrt 3 [/tex] ?
(1) [tex]\sqrt 2 [/tex] (2) [tex]3\sqrt 3 [/tex]
(3) Two are equal (4) None of the above
Ans. Now to prove this we subtract them. Now
[tex]\begin{array}{l}
\sqrt 2 - 3\sqrt 3 \\
= \frac{{\left( {\sqrt 2 - 3\sqrt 3 } \right)\left( {\sqrt 2 + 3\sqrt 3 } \right)}}{{\left( {\sqrt 2 + 3\sqrt 3 } \right)}}\\
= \frac{{2 - 9 \times 3}}{{\left( {\sqrt 2 + 3\sqrt 3 } \right)}}\\
= \frac{{2 - 27}}{{\left( {\sqrt 2 + 3\sqrt 3 } \right)}} < 0\left[ {as 27 > 2} \right]
\end{array}[/tex]
The greater number is [tex]3\sqrt 3 [/tex]
42. The fraction equivalent to [tex]\frac{2}{5}\% [/tex] is
(1) [tex]\frac{1}{{40}}[/tex] (2) [tex]\frac{1}{{125}}[/tex]
(3) [tex]\frac{1}{{250}}[/tex] (4) [tex]\frac{1}{{500}}[/tex]
Ans. [tex]\frac{2}{5}\% = \frac{2}{5} \times \frac{1}{{100}} = \frac{1}{{250}}[/tex]
43. If, [tex]\frac{1}{5}:\frac{1}{x}::\frac{1}{x}:\frac{1}{{1.25}}[/tex] , then x = ?
(1) 1.5 (2) 2
(3) 2.5 (4) 3.5
Ans.
[tex]\begin{array}{l}
\frac{1}{5}:\frac{1}{x}::\frac{1}{x}:\frac{1}{{1.25}}\\
\Rightarrow \frac{{{\textstyle{1 \over 5}}}}{{{\textstyle{1 \over x}}}} = \frac{{{\textstyle{1 \over x}}}}{{{\textstyle{1 \over {1.25}}}}}\\
\Rightarrow {\left( {\frac{1}{x}} \right)^2} = \frac{1}{5} \times \frac{1}{{1.25}}\\
\Rightarrow {\left( {\frac{1}{x}} \right)^2} = \frac{1}{{6.25}}\\
\Rightarrow \frac{1}{x} = \sqrt {\frac{{100}}{{625}}} = \sqrt {\frac{{10 \times 10}}{{25 \times 25}}} = \frac{{10}}{{25}}\\
\Rightarrow x = \frac{{25}}{{10}} = 2.5
\end{array}[/tex]
44. By selling 100 pencils, a shopkeeper gains the selling price of 20 pencils. His gain percentage is
(1) 25% (2) 20%
(3) 15% (4) 12%
Ans. Let selling price of 1 pencil is Rs. k .
Therefore the selling price of 100 pencils is Rs. 100k.
From the question the profit = selling price of 20 pencils = Rs. 20k.
Therefor the cost of 100 pencils is ( 100k - 20k ) = 80 k.
We know
[tex]\begin{array}{l}
Gain\% = \frac{{S.P - C.P}}{{C.P}} \times 100\\
= \frac{{100k - 80k}}{{80k}} \times 100\\
= \frac{{20k}}{{80k}} \times 100\\
= 25\%
\end{array}[/tex]
45. Two numbers are in the ratio of 3 : 5. If each number is increased by 10, the ratio becomes 5 : 7. The numbers are
(1) 3 , 5 (2) 7 , 9
(3) 13 , 22 (4) 15 , 25
Ans. Let two numbers are x and y.
From the question we get
[tex]\begin{array}{l}
x:y = 3:5 \Rightarrow \frac{x}{y} = \frac{3}{5}.........................(i)\\
\left( {x + 10} \right):\left( {y + 10} \right) = 5:7 \Rightarrow \frac{{\left( {x + 10} \right)}}{{\left( {y + 10} \right)}} = \frac{5}{7}.....................(ii)
\end{array}[/tex]
From (i) we get [tex]x = \frac{3}{5}y[/tex]................................(iii)
use (iii) in (i)
[tex]\begin{array}{l}
\frac{{\left( {x + 10} \right)}}{{\left( {y + 10} \right)}} = \frac{5}{7}\\
\Rightarrow \frac{{\left( {\frac{3}{5}y + 10} \right)}}{{\left( {y + 10} \right)}} = \frac{5}{7}\\
\Rightarrow 7\left( {\frac{3}{5}y + 10} \right) = 5\left( {y + 10} \right)\\
\Rightarrow 21y + 350 = 25y + 250\\
\Rightarrow 4y = 100\\
\Rightarrow y = 25
\end{array}[/tex]
Therefor [tex]x = \frac{3}{5} \times 25 = 15[/tex]
46. The simplified values of
[tex]\frac{1}{{56}} + \frac{1}{{72}} + \frac{1}{{90}} + \frac{1}{{110}} + \frac{1}{{132}}[/tex]
(1) [tex]\frac{1}{{84}}[/tex] (2) [tex]\frac{1}{{28}}[/tex]
(3) [tex]\frac{5}{{84}}[/tex] (4) [tex]\frac{1}{{12}}[/tex]
Ans.
[tex]\begin{array}{l}
\frac{1}{{56}} + \frac{1}{{72}} + \frac{1}{{90}} + \frac{1}{{110}} + \frac{1}{{132}}\\
= \frac{1}{{7 \times 8}} + \frac{1}{{8 \times 9}} + \frac{1}{{9 \times 10}} + \frac{1}{{10 \times 11}} + \frac{1}{{11 \times 12}}\\
= \left( {\frac{1}{7} - \frac{1}{8}} \right) + \left( {\frac{1}{8} - \frac{1}{9}} \right) + \left( {\frac{1}{9} - \frac{1}{{10}}} \right) + \left( {\frac{1}{{10}} - \frac{1}{{11}}} \right) + \left( {\frac{1}{{11}} - \frac{1}{{12}}} \right)\\
= \frac{1}{7} - \frac{1}{8} + \frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{12}}\\
= \frac{1}{7} - \frac{1}{{12}}\\
= \frac{{12 - 7}}{{84}} = \frac{5}{{84}}
\end{array}[/tex]
