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]
