Quantitative Aptitude Test - 8

Let’s say Dev’s share is Rs 's'.

Then Suma gets Rs (250000 – s)

Interest received by Dev compounded at 8% for 2 years:

= s[1 + 8/100]2 – s

= (27/25)2 s – s

= 104 s / 625

Interest received by Suma at 10% per annum simple interest would be

=(250000 – s) x 10 x 2/100

= [(250000 – s)/5(250000 – s)/5]

= 104 s / 625 + 13,360

Implies 625 (250000 – s) = 5 (104s + 625 x 13360);

or s = Rs 1,00,000

2,50,000 -1,00,000

Hence Suma’s share = Rs 1,50,000. Option D is correct.

If O is fixed in the middle then there are four places left to be filled by the remaining four alphabets (A, B, V, E).

This can be done in 4! Ways. Thus, the total number of ways = 4!

= 24

Option D is correct.

Exactly one of ab, bc and ca is odd => Two are odd and one is even.

abc is a multiple of 4 => the even number is a multiple of 4.

The arithmetic mean of a and b is an integer => a and b are odd.

and so is the arithmetic mean of a, b and c. => a + b + c is a multiple of 3.

c can be 4 or 8.

c = 4; a, b can be 3, 5 or 5, 9

c = 8; a, b can be 3, 7 or 7, 9

Four triplets are possible.

Let the son's present age be x years.

Then, according to question (38 - x) = x

2x = 38.

x = 19.

Son's age 5 years back (19 - 5) = 14 years.

Let the free luggage allowance be ‘f’ kg. Let the weight of the luggage carried by Ravind be ‘r’ kg and the weight of the luggage carried by Pranas be ‘p’ kg. Thus, the excess luggage weights carried by Ravind and Pranas respectively are (r – f) kg and (p – f) kg.

Thus, the total luggage charge for both would be (r – f)k + (p – f)k if k is the charge per kg.

Thus, (r – f)k + (p – f)k = 1100.

(r + p – 2f)k = 1100 .......................(1)

If Ravind carried twice the luggage weight he actually did, i.e., if he carried 2r kg, then the excess luggage weight he carried would have been 2r – f and the corresponding charge would have been (2r – f) k.

Therefore, (2r – f)k = 2000 ................(2)

Likewise, If Pranas carried twice the luggage he actually did i.e., if he carried 2p kg, then the excess luggage he carried would have been 2p –f and the corresponding charge would have been (2p – f) k.

Therefore, (2p – f)k = 1000 ................(3)

Adding (2) and (3) and simplifying, we get,

(r + p – f)k = 1500 ........................(4)

Dividing (4) by (1) and simplifying, we get,

19f = 4r + 4p .............................(5)

Dividing (2) by (3) and simplifying, we get,

–f = 2r – 4p .............................(6)

Solving (5) and (6) for r, we get,

r = 3f ...................................(7)

Subtracting (1) from (4) and simplifying, we get,

fk = 400.

Ravind’s luggage charge = (r – f)k.

But, according to equation (7), r = 3f.

Therefore, Ravind’s luggage charge = 2fk

But, fk = 400.

Therefore, Ravind’s luggage charge = Rs. 800.

Hence, the answer is "Rs. 800".

Total surface area of the box = 2(4 * 6 + 1 * 6 + 1 * 4)

= 2(24 + 6 + 4)

= 68 cm²

As the problem says, paper can’t be torn/cut a portion of paper will need to be fold, so, the area of paper required would be greater than 68 cm².

Only option b) gives the area greater 68 cm²

Hence, the answer is 12 cm by 6 cm

The result has to be multiple of 137 and 173.

Checking option-wise:137 × 173

= 23,701137 × 173 × 2

= 47,402137 × 173 × 3

= 71,103137 × 173 × 4

= 94,804137 × 173 × 8

= 1,89,608

Choosing 23,701, each front wheel will take 173 revolutions and each rear wheel will take 137 revolutions.

Difference = 173 – 137 = 36

Choosing 47,402, each front wheel will take 346 revolutions and each rear wheel will take 274 revolutions.

Difference = 72

Choosing 71,103, each front wheel will take 519 revolutions and each rear wheel will take 411 revolutions.

Difference = 108

Therefore, the correct option is C.

Let the length of Train A be x meters

Let speed of Train B be y kmph

Relative distance = Relative speed * time taken to cross/overtake

Crossing scenario:

Relative speed of 2 trains = 63 + y

Time taken to cross = 27 sec or 27/3600 hrs

Relative distance between 2 trains = Length of Train A + length of train B = (x + 0.5) km

Therefore, x + 0.5 = (63 + y) * 27 / 3600 ----- (1)

Overtaking scenario:

Relative speed of 2 trains = 63 – y

Time taken to overtake = 162 sec or 162/3600 hrs

Relative distance between 2 trains = x + 0.5

Therefore, x + 0.5 = (63 – y) * 162/3600 --- (2)

From (1) and (2), solve for y.

(63 + y) * 27 = (63 – y) * 162

27y + 162 y = 63*162 – 63 *27

189y = 63 * 135 or y = 45 kmph

Substitute in (2) to get x.

x + 0.5 = (63 – 45) * 162/3600

Or x = 0.31 km or 310 meters

Speed = 54 x$\frac{5}{18}m/sec$

=15 m/sec

Length of the train = (15 x 20)m = 300 m.

Let the length of the platform be x metres.

then,$\frac{X+300}{36}=15$

x + 300 = 540

x = 240 m.

According to the statement:-

=> X = YQ + 25

So, 3 times of X = 3YQ + 75

If 3 times of X is divided by Y gives a remainder 4

Then (3YQ + 75) / Y gives a remainder 43YQ is divisible by Y and only when 75 is divided by 71 gives a remainder 4. Therefore the value of Y is 71.

Hence, option D is correct.