capegemini written interview questions and answers paper 409 - skillgunNote: Paper virtual numbers may be different from actual paper numbers . In the page numbers section website displaying virtual numbers .
The rectangular shaped building dimensions are P x Q. If the breadth is increased by 40% and the length is decreased by 30% then the % of area of the new building compared with previous one is:
Dimensions of the building = P x Q. Assume that, P is length and Q is breadth.
Area of original wall = PQ unit2.
New length (after decreasing 30%) = P– 30P/100 = 70P/100
New breadth (after increasing 40%) = P = P + 40P/100 = 140P/100.
New area = 70P/100 x 140P/100 = 49PQ/50.
Therefore, required % = (new area / original area ) x 100 = ((49PQ/50) / PQ) x 100= 98%.
If the square shaped place area decreases by 36%, then the % of each side of the place decreases by:
Assume that the original area of the place is 100 unit2.
Then, its original sides = sqrt(100) = 10 units.
% decrease in area = 36%
New area = 100 - 36 = 64 unit2.
new side of the place = sqrt(64) = 8 units.
i.e decrease on 10 units = (10 - 8) = 2 units.
Decreasing % = (units of decreasing / original units) x 100 = 2 x 100/10 = 20%
The dimension of a plot is L m x B m. If l and b decreases by 20% and 40% respectively, then the area of the plot before alteration exceeds the area of new one by:
Assume length = L m and breadth = B m.
Original area = (LB) m2.
New length (after decreasing 20%) = (100 - 20)% of L= 80% of L = 80L /100 m = 4L / 5 m.
New breadth (after decreasing 40%) = (100 - 40)% of B = 60% of B = 60B / 100 = 3B / 5m.
New area = 4L/5 x 3B/5 = 12(LB)/25 m2.
Difference in area = original area - new area = LB - 12(LB)/25 = 13(LB)/25 m2.
Therefore, required % = difference / original area x 100 = [(13(LB)/25) / (LB) ] x 100 = 13 x 4 = 52%.
Two cans A and B contains milk worth Rs.7 per litre and Rs.9 per litre respectively. If the contents of A and B are transferred to another can C in the ratio 3 : 7 then the cost per litre of the mixture in can C is:
Cost of 1 litre of A = Rs.7 = cheaper quantity.
Cost of 1 litre of B = Rs.9 = dearer quantity.
Let the mean price be Rs.x
Applying the rule of alligation,Therefore, (Cheaper quantity) (Dearer quantity) = (d - m) : (m - c) = (9 -x ) : (x - 7)
Given ratio = 3 / 7 = 9 - x/ x- 7
63 – 7x = 3x - 21
x = 8.4.
Two qualities of rice at Rs.63 per kg and Rs.67.50 per kg are mixed with another quality of rice in the ratio 2:2:3. The final mixture sold at Rs.76.50 per kg then the rate of third quality rice per kg was:
Given that, first and 2nd varieties are mixed in equal proportion.
Their average price = Rs.(63 + 67.50) / 2 = Rs.65.25
The new mixture is formed by mixing two varieties, one at Rs.65.25 per kg and the other at Rs.X per kg in the ratio 4:3. (note that, 2:2:3 becomes 4:3).
Given that, mean price of new mixture = Rs.76.50
We have to find X.
Applying the rule of alligation,Therefore, (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c) =
X - 76.50 / 11.25 = 4/3
3X – 229.50 = 45
X = 91.50
Hence, the required price of 3rd quality rice = Rs.91.50.
In what ratio, a liquid A of cost Rs.31 per litre should be mixed with liquid B of cost Rs.36 per litre, so that the cost of the liquid of mixture is Rs.32.35 per litre?
Cost of 1 litre of liquid A = Rs.31 = cost of cheaper quantity.
Cost of 1 litre of liquid B = Rs.36 = cost of dearer quantity.
Given, mean price = Rs. 32.25
Applying the rule of alligation
Therefore, required ratio = (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c)
= 3.75 : 1.25 = 3:1.
A fruit seller has 16 apples and cost of each apple is Rs.12. Three persons A,B and C decided to buy that apples. If A and B gives Rs.84 and Rs.48 to the seller for some apples, then how many apples C can buy?
Given that the seller has 16 apples and he sells each apple at Rs.12
Then total cost of 16 apples = 16 x 12 = Rs.192
A and B buys for Rs.84 and Rs.48.
Then the cost of remaining apples = Rs.(192 - (84+48))
TC can buy for Rs.60
Since each apple costs Rs.12, then C can buy 60/12 = 5 apples.
Hence the answer is 5.
Rajesh bought P number of sports goods for Rs.9000. If each item was cheaper by Rs.30 then with the same amount he could have bought 50 more items than P. Find the number of items bought by Rajesh ?
Rajesh have bought P items for Rs.9000.
Then the cost of each item = Rs.9000/P If each item was cheaper by Rs.30, he could have bought 50 more items.
Therefore, with Rs.30 discount, the amount of each item = Rs.9000 / X + 50 ...(2)
he bought 50 more items when cost of each item is less by Rs.30 then we have
9000/P - 9000/P + 50 = 30
1/P - 1/P + 50 = 30/9000
50 / P(P + 50) = 1/300
P(P + 50) = 15000
P2 + 50P = 15000
P2 + 50P - 15000 = 0
(P+150)(P-100) = 0
P = -150 or P = 100
We don't take negitive value P = 100
On Independence day, chocolates were to be distributed among 350 kids in a play school. But 140 kids were absent on that day and each kid present got 3 chocolates more. Find the total number of chocolates bought for the distribution.
Assume that total number of chocolates be Q.
Total number of kids in the school(when everyone is present) = 350.
Then number of chocolates per kid = Q/350 chocolates ...(1)
But,140 kids are absent.
Total number of kids present = 350 - 140 = 210
Number of chocolates per present kid = Q/210 chocolates ...(2)
210 kids were present and each kid gets 3 extra.
Or Q/210 - Q/350=3
350Q - 210Q / (210 x 350)=3
140Q/ 210 x 350=3
Q = 3 x 210 x 350 / 140 = 1575
8 friends planned to go to hotel for dinner and to share the bill amount equally. If one of them have forget to bring the wallet, then what will be the extra amount contribute by each to pay the bill of Rs.1904 ?
Given that the bill amount = Rs.1904
Actual share of each = Rs.1904/8 = Rs.238
If one of 8 is left, then the sharing amount = Rs.1904/7 = Rs.272
Then the extra amount given by each = Rs.(272 - 238) = Rs.34.
Hence the answer is Rs.34 ...
The Score's ratio of two teams P and Q is 5:8. If the score of team P is increased by 60% and those of Q decrease by 35%, then the new ratio of their score becomes 9:8.Find the score of team P.
Assume that original score of P and Q be 5x and 8x respectively.
New score of P after increasing 60% = 160% of 5x
160 / 100 x 5x = 8x
New score of Q after decreasing 35% = 65% of 8x
65 / 100 x 8x = 26x / 5
Therefore 8x : 26x / 5 = 9 : 8 (given)
40x / 26x = 9 / 8
This cannot give x value
Hence the given data is inadequate
805 ml of mixture1 is mixed with 700ml of mixture2. If mixture1 has acid1 and acid2 in the ratio 4:3 and mixture2 has acid2 and acid3 in the ratio 2:5 then the amount of acid2 in the new mixture is:
acid1 and acid2 is in the ratio 4:3, then
Quantity of acid2 in 805ml of mixture1 = 805 x 3 / 7 = 345 ml
And acid2 and acid3 in the ratio 2:5.
Quantity of acid2 in 700ml of mixture2 = 700 x 2 / 7 = 100 x 2 = 200ml
Therefore quantity of acid2 in new mixture = 345 + 200 = 545ml.
Hence the required answer is 545 ml.
A Sugar solution of 2430 ml is poured into 3 cans namely P,Q and R such that 5 ml,10 ml and 15 ml are taken out and the remaining solution in P, Q and R will be in the ratio 3 : 4 : 5. Find the initial amount of solution in Q.
Total amount of solution = 2430 ml
After taking 5 ml,10 ml and 15 ml out, the remaining amount of Solution = 2430 - (5 + 10 + 15) ml
= 2400 ml
Therefore 2400 ml of solution is in the ratio = 3 : 4 : 5
Amount of solution present in Q now = 4/12 x 2400 = 800 ml
Initial amount of solution in bottle Q = New amount of solution in Q + Solution taken out from Q
i.e., 800 + 10 = 810 ml contained in bottle Q initially.
The ratio of length of sides of a machine is 1/6:1/4:1/3 and the sum of length of sides is 104, If we increase each side by 3 m then the length of largest side will be:
The sides of machine are in the ratio 1/6 : 1/4 : 1/3 = 4:6:8
Assume the sides of the machine be 4x,6x and 8x
Given that the sum of the sides = 104 m
i.e., 4x + 6x + 8x = 104 m
x = 104/18 = 52/9
Then the largest side 8x = 8 x 52/9 = 416 / 9
The sides are increased by 3 m.
Now the length of the largest side = 416 / 9 + 3 = 443 / 9
= 49.22 = 49 m
The sum of the 1/2 of Karunanidhi's age two years from now and 1/3 of his age three years ago is twenty years, then how old is he now?
Karunanidhi's age two years in the future and three years in the past.
Assume Kaunanidhi's age now be K.
His age after two years from now is K + 2 and three years ago is K - 3.
1/2 of age two years from now is (1/2)(K + 2) = K/2 + 1
1/3 of age three years ago is (1/3)(K - 3) = K/3 - 1
The sum of these two numbers is twenty,
i.e.k/2 + 1 + K/3 - 1 = 20
K/2 + K/3 = 20
3K + 2K = 120
K = 120/5 = 24
Jagan is 3 years elder than Raja. Naga was 28 years of age when Mahesh was born while Sahitya was 26 years of age when Raja was born. If Mahesh was 4 years of age when Jagan was born, then what was the age of Naga and Sahitya respectively when Jagan was born ?
32 &23 years
42 & 24 years
23 & 28 years
24 & 32 years
Given that, " Naga was 28 years of age when mahesh was born" If Mahesh was 4 years of age then Naga's age is 28 + 4 = 32 years.If mahesh was 4 years of age when Jagan was born.
From the above two statements we would have Naga's age when Jagan was born is 32 years.
Also given that, Jagan is 3 years elder than Raja and sahitya was 26 years of age when Raja was born.
Sahitya's age when Jagan was born = 26 - 3 = 23 years
My father was 28 years older than my brother and my mother was 23 years elder than my sister when I was born. If my brother is 10 years older than me and my mother is 4 years younger than my father, how old was my sister when I was born?
When I was born, my brother was 10 year old,
My father's age is 10 + 28 = 38 yrs. old
My Mother's age is 38 - 4 = 34 years old and my sister's age is 34 - 23 = 11 year old
Hence the answer is 11 years.
The two digit number which is there in set that is selected at random so that, we have to find the probability that the selected number will be a multiple of '8'.
There are a total of 90 two digit numbers.
So,the set contains 90 numbers.
We know that the first two digit number that is a multiple of 8 is 16 and every 8th number from 17 will be a multiple of 8. there are 11 (integral value of 90/8 = 11) of those numbers that are divisible by '8'.(alternatively the two digit multiples of 8 are 16,24,32,40,48,56,64,72,80,88 and 96)
Then, the required probability is 8/90 = 4/45
Find the probability that a number selected at random from the set of 3 digit numbers will be a multiple of '9' and a divisor of 900?
The three digit numbers are 100,101,102,...,999
There are a total of 900 three digit numbers.
We know that the number 99 preceded by 100 is divisible by 9 and then every nine-th number from 101 will be divisible by '9'.
Therefore, there are (900/9) 100 of those numbers that are divisible by '9'.
Now, the 3 digit divisors of 900 are 900/1, 900/2, 900/3, 900/4, 900/5, 900/6 and 900/9.
i.e., the 3 digit divisors of 900 are 900, 450, 300, 225, 180, 150 and 100.
Here, except 150 and 100 other numbers are multiple of 9.
Find the probability that a two digit number selected at random will be a divisor of '96' and not a divisor of 24?
There are totally 90 two digit numbers.
The divisors of 96 are 1,2,3,4,6,8,12,16,24,32,48 and 96.
The two-digit divisors of 96 are 12,16,24,32,48 and 96.
Here, the two digit divisors of 24 are 12 and 24 only.
Therefore, 16,32,48 and 96 are the required numbers.
Hence the required probability = 4/90 = 2/45
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