# TCS Ninja, Digital, NQT Placement Questions with solutions - 18

1. 2ab5 is a four digit number divisible by 25. If a number formed from the two digits ab is a multiple of 13, then ab is
a. 52
b. 45
c.10
d.25
Sol: For a number to be divisible by 25, last two digits of that number should be divisible by 25. So b must be either 2 or 7
it is given that ab must be divisible by 13 and in the options only 52 is divisible by 13.

2. The average temperature of Tuesday Wednesday and Thursday was 37 C. The average temperature of Wednesday and Thursday and Friday was 38 C. if the temperature on Friday was 39 C.
Find the temperature on Tuesday.
a. 37.33
b. 38.33
c. 36
d. None of the above
Sol:
(Tues + Wed + Thurs)/3=37
Tues + Wed + Thurs=111...(1)
(Wed + Thurs + Fri)/3=38
(Wed + Thurs + Fri) =114...(2)
Given friday is 39.
Then, (2) - (1) Fri - Tues = 3
So 39 - Tues = 3
Tuesday =36

3. There are 5 boxes in a cargo. The weight of the 1st box is 200 KG, the weight of the 2nd box is 20% higher than the third box, whose weight is 25% higher than the 1st box weight. The 4th box which weighs 350 KG is 30% lighter than the 5th box. Find the difference in average weight of the 4 heaviest boxes and the four lightest boxes.
Sol: weight of 1st box=200
weight of 3rd box=(125/100)*200=250
weight of 2nd box=(120/100)*250=300
weight of 4th box =350
weight of 5th box=(10/7)*350=500
average of 4 highest weighted boxes=(500+350+300+250)/4=350
average of 4 lightest boxes=(350+300+250+200)/4=275
therefore difference=350-275=75

4. The length, breadth and height of a room are in the ratio 3:2:1. If the breadth and height are halved, while the length is doubled. Then the total area of the 4 walls of the room will be decreased by
a. 30%
b. 18.75%
c. 15%
d. 13.6%
Sol: Given l:b:h=3:2:1
let h=10, b = 20, and l = 30
area = $2(l + b)h$
area= 2*(3x+2x)*x = $2(30 + 20)10 = 1000$
Now after those adjustments in the measurements,
l=60, b=10, h=5
area= $2(l + b)h$ = $2(60 + 10)5 = 700$
Percentage decrease= $\displaystyle\frac{{1000 - 700}}{{1000}} \times 1000 = 30\%$

5. A circle circumscribes three unit circles that touch each other. What is the area of the larger circle? Note that p is the ratio of the circumference to the diameter of a circle ( 3.14159265).
Sol:

By joining centers of 3 unit circles we will get an equilateral triangle of length 2 unit. We have to find the length of the orange line.
And center of the equilateral triangle will be the center of the big circle.
So radius of the big circle will be = (1 + Circum radius of the equilateral triagle)

Formula for Circul radius of the equilateral triangle = $\displaystyle\frac{2}{3} \times \left( {\frac{{\sqrt 3 }}{2}a} \right)$ here ${\displaystyle\frac{{\sqrt 3 }}{2}a}$ is the height of the triangle.  a is the side of the triangle
Circum radius of equilateral triangle = $\displaystyle\frac{2}{3} \times \displaystyle\frac{{\sqrt 3 }}{2} \times 2 = \displaystyle\frac{2}{{\sqrt 3 }}$

Area of big circle will be =$\pi {r^2} = 3.14 \times {(1 + \displaystyle\frac{2}{{\sqrt 3 }})^2}$ = $3.14 \times \left( {1 + \displaystyle\frac{4}{{\sqrt 3 }} + \frac{4}{3}} \right)$
$= 3.14 \times \left( {1 + \displaystyle\frac{4}{{\sqrt 3 }} + \frac{4}{3}} \right)$ = $3.14 \times \left( {\displaystyle\frac{7}{3} + \displaystyle\frac{4}{{\sqrt 3 }}} \right)$
$= 3.14 \times \left( {\displaystyle\frac{{7 + 4\sqrt 3 }}{3}} \right)$

6. Rajesh calculated his average over the last 24 tests and found it to be 76. He finds out that the marks for three tests have been inverted by mistake. The correct marks for these tests are 87, 79 and 98. What is the approximate percentage difference between his actual average and his incorrect average?
Sol: No Change
Incorrect value is: 78, 97, 89
correct values are: 87, 79, 98
difference between correct and incorrect value is= 9 + 9 -18=0

7. Joke is faster than Paul, Joke and Paul each walk 24 KM. The sum of their speed is 7 Km per hour. And the sum of times taken by them is 14 hours. Then, Joke speed is
a. 3 KM/Hr
b. 4 KM/Hr
c. 5 KM/Hr
d.7 KM/Hr
Sol:
$Speed = \displaystyle\frac{{Distance}}{{{\rm{Time}}}}$
let the speed of joke x then speed of paul will be 7-x
$\displaystyle\frac{{24}}{x} + \displaystyle\frac{{24}}{{7 - x}} = 14$
Try to plugin the values from the options. If Joke speed is 4 the paul is 3.

8. The crew of a rowing team of 8 members is to be chosen from 12 men (M1, M2, …., M12) and 8 women (W1, W2,…., W8), such that there are two rows, each row occupying one the two sides of the boat and that each side must have 4 members including at least one women. Further it is also known W1 and M7 must be selected for one of its sides while M2, M3 and M10 must be selected for other side. What is the number of ways in which rowing team can be arranged.
SoL:
We need two person for one side and 1 women for the another side.  We select that women in 7 ways. Now that second side people can sit in 7x4! ways.
Now for the first side we need two people from the remaining 14. So this can be done in ${}^{14}{C_2}$ ways and this side people can sit in ${}^{4}{C_2} \times 4!$ ways.
Again the first group may take any of the two sides.  So total ways are $2 \times 7 \times 4! \times {}^{14}{C_2} \times 4!$

9. In a certain city, 60% of the registered voters are congress supporters and the rest are BJP supporters. In an assembly election, if 75% of the registered congress supporters and 20% of the registered BJP supporters are expected to vote for candidate A, what percent of the registered voters are expected to vote for candidate A?
Sol: let the people in the city be 100
Congress supporters = 60% of 100 = 60
40% are BJP=40% of 100 = 40
out of 60,75% voted for congress=75%(60)=45
out of 40%,20% voted for congress=20%(40)=8
Total=45 + 8 = 53
Total percent= 53%

10. Anusha, Banu and Esha run a running race of 100 meters. Anusha is the fastest followed by Banu and then Esha. Anusha, Banu and Esha maintain constant speeds during the entire race. When Anusha reached the goal post, Banu was 10m behind. When Banu reached the goal post Esha was 10m behind. How far was behind Anusha when the latter reached the goal post.
option
a) 70
b) 81
c) 90
d) 80
Sol:
By that time Anusha covered 100m, Bhanu covered 90m. So ratio of their speeds = 10 : 9
By that time Bhanu reached 100m, Esha covered 90m. So ratio of their speeds = 10 : 9
Ratio of the speed of all the three = 100 :  90 : 81
By that time Anusha covered 100m, Esha Covers only 81.

11. Seven different objects must be divided among three persons. In how many ways this can be done if at least one of them gets exactly one object.
Sol: Division of m+n+p objects into three groups is given by $\displaystyle\frac{{(m + n + p)!}}{{m! \times n! \times p!}}$
But 7 = 1 + 3 + 3 or 1 + 2 + 4 or 1 + 1 + 5
So The number of ways are $\displaystyle\frac{{(7)!}}{{1! \times 3! \times 3!}} \times \frac{1}{{2!}}$ + $\displaystyle\frac{{(7)!}}{{1! \times 2! \times 4!}}$ + $\displaystyle\frac{{(7)!}}{{1! \times 1! \times 5!}} \times \frac{1}{{2!}}$ = 70 + 105 + 21 = 196

12. George while driving along the highway saw road markers which are at equal distances from each other. He crosses the markers every 20 seconds. If he increases his speed by x meters per second, he crosses the markers at every 15 seconds. But if he increases his speed by y meters per second, he crosses the marker at every 10th second. If y-x = 40 meters per second, then what is the distance between two markers.
Sol: Let speed be =z m/s then Distance= 20z m
(z+x)15=20z; (z+y)10=20z
Also given that y - x = 40
solving we get 20z=1200

13. How many different 9 digit numbers can be formed from the number 223355888 by re-arranging its digits so that the odd digits occupy even position?
Sol: Odd places are 4 and these are occupied by 3355. So this can be done in 4!/ (2! 2!) = 6
There are 5 even numbers which have to be placed at 5 odd places. So 5!/(2!3!) = 10 ways
so total number of ways of arranging all these numbers are 10 * 6 = 60 ways

14. In a vessel, there are 10 litres of alcohol. An operation is defined as taking out five litres of what is present in the vessel and adding 10 litres of pure water to it. What is the ratio of alcohol to water after two operations?
a) 1 : 5
b) 2 : 3
c) 1 : 6
d) 3 : 2
Sol: Final concentration = Initial concentration $\left( {1 - \displaystyle\frac{\text{replacement quantity}}{\text{final volume}}} \right)$
Final concentration = ${\rm{ = 1}} \times \left( {{\rm{1 - }}\displaystyle\frac{{10}}{{15}}} \right) = \displaystyle\frac{1}{3}$

Final concentration = $\displaystyle\frac{1}{3} \times \left( {{\rm{1 - }}\frac{{10}}{{20}}} \right) = \displaystyle\frac{1}{6}$
So ratio of alcohol : water = 1 : 5