# Cocubes Questions Quants from Previous Year Questions 11

201. In how many ways can 34 people be divided into 17 couples?
a. (34!)/{(17!)17 (2!)}    b. (34!)/{(2!)17 (17!)}   c. (34!)/ {(2!)(17!)}   d. Data Inadequate
Ans:  Short cut how many ways n people be divided into n/2 couples
(n!)/{(2!)n/2 (n/2)!}  so ans is b. (34!)/{(2!)17 (17!)}

202. The electricity bill of JMD is partly fixed and partly varies as the number of units of electricity consumed. When is a certain month 650 units consumed the bill was Rs. 2,130. In yet another month 720 units were consumed and the bill was Rs. 2,340. What would be the bill for the month 940 units consumed?
a. 3,575   b. 3,000   c. 4,350   d. 2,990
Ans: 3,000
Solution: let’s say fixed  cost is k and unit cost is x
Then k+650 * x=2130
K+720 * x= 2340
From above two equations k=180 and x=3
Now 180+940*3= 3000

203. A wooden board is 7ft 9 inches long. It is divided into 3 equal parts. What is the length of each part?
a. 2 ft 7 inches   b. 3 ft    c. 2 ft 4 inches    d. 2ft 6 inches

Ans: 2 ft 7 inches
1 ft= 12 inches
So 7 ft 9 inches = 93 inches
Divide 93 inches into 3 equal parts. Then each part is 31 inches.
31 inches = 2 ft 7 inches
204. Determine the metal required to make a 21 m long pipe if its inner and outer diameter are 12 m and 10 m respectively.
a. 2904 m3  b. 2534 m3    c.  2843 m3    d. 2647 m3
volume of a hollow cylinder=pi*h*(R^2-r^2)
22/7*21m*(144-100)m^2
=66*44m3
=2904m3

205. Two trains of length 180 m and 220 m are running in opposite directions, the first one at the rate of 50 kmph and while second one at       . Calculate the time they will require to pass each other.
a. 20 seconds   b. 16 seconds   c. 18 seconds   d. 17 seconds

206.  30 men can do a work in 30 days and 40 women can do the same work in 40 days. If they started working together, how many more men required completing the work in 10 days?
a. 42   b. 28   c. 25   d. 38
Ans:  38
Total work has to be finished in 10 days. That means 1 day work becomes 1/10
Let’s say x number of men added to complete the work in 10 days.
Then
I day’s work of ( 30 men + 40 women + x men) = 1/10    —- (1)
If 30 men can do a work in 30 days
Then 30 men 1 day’s work = 1/30
Then 1 man 1 day’s work = 1/900
40 women can do a work in 40 days
Then 40 women 1 day’s work = 1/40
Substitute above values in equation (1)
(1/30  + 1/40 + x (1/900))=1/10
Simplify above equation and find x value.
X=37.5  but x cannot be decimal value because x represents number of men. So x=38

207.   Read the information given below and answer the question that follows:
I. (x & y) = ( x2 – y2)
II. (x ? y) = (x-y)/2
III. (x \$ y)= (x + y)/2

If (x @ y) is defined as (x3-y3) then for integers x,y>2 and x > y which of the following relationships will always be true?
a.       (x & y) < (x @ y)   b. (x\$ y)  > (x?y)    c. (x \$ y) ≥ (x ?y)   d.  Both (x\$ y)  > (x?y)     and (x & y) < (x @ y)

208. 3 designers x,y and z can stitch 324 dresses in 6 weeks working simultaneously. During one shift, z stitches as many dresses more than y as y stitches more than x. Z’s work in 10 weeks in equivalent to x’s work in 14 weeks. How many dresses does x stitch per shift.
a. 21  b 15  c. 27  d. 18

209. A person travels for 3 hours at the speed of 40 kmph and for 4.5 hours at the speed of 60 kmp. At the end of it, he finds that he has covered (3/5)th  of the total distance. At what average speed should he travel to cover the remaining distance in 4 hours?
a. 70 kmph   b. 65 kmph   c. 75 kmph  d. 60 kmph
Ans:  65 kmph
Distance covered in 3 hours = 3 * 40  = 120 km
Distance covered in 4.5 hours = 60 * 4.5 = 270 km
Let’s say total distance is x kms
Then (3/5) * x= 120+270
X= 650 kms
Remaining distance = 650 – (120+270) = 260 km
Speed = distance / time = 260 / 4 = 65 kmph

210. If the simple interest is 7% annual and compound interest is 6% annual, find the difference between the interests after 4 years on a sum of Rs 2000.
a. 33.05    b. 32.5    c.   37.5    d. 35.05
Ans: 35.05
Simple interest after 4 years = (2000 * 7 * 4)/100= 560
Compound interest after 4 years = 2000*( 1+ 6/100)4 – 2000 = 525.95
Difference = 560 – 525.95 = 35.05

211.  if (p+q)=3 then what is the value of (p3 + q3 ), when it is given that p=1/q?
a. 123   b. 143   c. 111  d. 132
Ans: 18
212. let M = (1,2,3,4, …. ) be the set of natural numbers and Q(n) be a mathematical statement involving the natural numbers “n” belonging to M such that
I. Q(1) is true i.e., Q(n) is true for n=1
II. Q(n+1) is true whenever Q(n) is true
Which of the following statements is true with regard to the given information?
a. Q(n) is true for all natural numbers n     b. Cannot be detrmined from the data given
c.  Both the mentioned statements are true.  D. Q(n) is true implies that Q(n+1) is true.

213. 132 ml of a drink contains vodka and water in 27:6. How much more water is to be added to get a new mixture
a. 24 ml  b. 54 ml    c. 12 ml   d. 36 ml

214. if (x + 1/x)= 3 then find the value of (x8+1/x8)
a. 3037   b. 2207   c. 1000   d. 800

215. The sum of ages of two people is 40 years. After 5 years the ratio of their ages will be 3:7 what is the ratio of their present ages?
a. 1:3   b. 1:2  c. 2:1   d. 3:1
Ans: a+b=40  — (1)
(a+5)/(b+5) = 3/7   — (2)
Solve above two equations then a=10 and b=30
So ratio is 1:3
216. Two numbers are in the ratio 11:6 if the first number is increased by 200 and second by 50% then the new ratio becomes 5:3 determine the original numbers
a. 40 and 25   b. 24 and 15  c. 550 and 300   d. 32 and 20
Ans: original ratio = 11k/6k
(11k+200) / (6k+3k)  = 5/3
Then k=50

So original numbers are 550 and 300
217. A photograph is to be fitted in a photo frame of sides 18 cm by 15 cm such that there is a margin of 1.5 cm left. What should be the area of the photograph?
a. 140 cm2  b. 180 cm2    c.  None of these   d. 270 cm2
Ans: 180  ( similar to q.no 78 )
1.5 cm margin on bothsides….then 1.5+1.5=3cm
area of photograph(rectangle) =(18-3)*(15-3)
=180cm2
218. if m+ (m-3)-1 = 6  then determine the value of (m-3)2 + (m-3)-2
a. 3  b -1  c. 1  d. 0

219. a peacock is sitting on a 19 m long pole, a snake is approaching the hole which is at bottom of the pole, the snake is 27 m away from the hole, if their speeds are same, find the distance from the hole at which the peacock pounces over the snake.
a. 3.4 m   b. 6.8 m   c. 5.9 m   d. 7.3 m
They both have same speed hence if snake moves ‘x’ then peacock also moves ‘x’ but diagonally down from the pole…
hence path of peacock , ground, and the pole make a triangle…… with diagonal (Hypotenuse) = x
snake moves ‘x’ hence distance left =( 27 – x ) which is the base of triangle
and height of the pole is the height of triangle =19.
now, apply pythagoras’ theorem—– x^2= (27-x)^2 + 19^2
solve the equation ….. x= 20.18 m
And the distance from the hole at which the peacock pounces over the snake = 27 – 20.18 = 6.8

220. At the bottom of the tank containing 100 gallons of water, a leak was formed of diameter “D” cm that can empty 5 gallon/minute. Due to the load of water the diameter (D) of the leak started varying every minute with respect to (t) as “Dt” cm. In what time the whole tank will become empty?

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