Quantitative Ability (S.A.) Test - 5

Here x lies between −7 and 13, then both (13 − x) and (x + 7) are positive.

Now (13 − x)⁷ (x + 7)³ is the greatest when

equal to (13−x)/7 and 3 factors each equal to (x+7)/3

The product of factors has maximum value, when sum of the factors is constant and each factor is equal

Here, sum of the factors

We can have the following possible combinations of the unit's digit of the three 3 -digit numbers whose sum is 9 or 19

1. (0, 0 and 9) : Last digit of the product is 0

2. (1, 9 and 9) : Last digit of the product is 1

3. (1, 2 and 6) : Last digit of the product is 2

4. (2, 3 and 4) : Last digit of the product is 4

5. (1, 3 and 5) : Last digit of the product is 5

6. (1, 4 and 4) : Last digit of the product is 6.

7. (1, 1 and 7) : Last digit of the product is 7.

8. (3, 7 and 9) : Last digit of the product is 9..

In total there are 8 possible digits which can be the last digit of the product of three 3 -digit numbers.

So there are only two digits (3 and 8), which are not possible.

Let the third proportion be x.

So, 27 : 36 ∷ 36 : x

∴ 27 × x = 36 × 36

⇒ x = 48

Hence, the correct option is (A).

To find the digits at the ten's and the unit's place we need to find the remainder when the number is divided by 100.

When 808¹¹¹¹ is divided by 100.

808¹¹¹¹ = (800 + 8)¹¹¹¹ since 800 is an integral

multiple of 100 the remainder is when 8¹¹¹¹ is divided by 100.

Remainder = 25 − 2 = 23

Actual remainder = 23 × 4 = 92

When 806¹¹¹¹ is divided by 100.

806¹¹¹¹ = (800 + 6)¹¹¹¹. since 800 is an integral multiple of 100 the remaincler is when 6¹¹¹¹ is divided by 100

When 216 is divided by 100, remainder = 16 So, when 216.4²⁵² is divided by 100 , remainder will be when 16 × 16 = 256 is divided by 100, i.e. 56

Therefore the net remainder when (92+56) is divided by 100, i.e. 48

Sum of the digits at the ten's and the unit's place

= 4 +8 = 12

||x − 2| − 1| < 6 ⇒ − 6 < |x − 2| − 1 < 6

⇒ |x − 2| <7

⇒ −7

∴ -5 since x is an integer, x can only assume integral values from -4 to 8.

Now, |y − 1| −2| < 9

⇒ −7 < |y−1 | < 11

⇒ |y − 1| < 11

⇒ −11 ∴ -10 since y is an integer, y can only assume integral values from -9 to 11.

Hence, −26 ≤ (x − 2y) ≤ 26

In an improper fraction, the numerator is greater than or equal to the denominator and denominator should not be equal to zero.

e.g. 61/5, 6/6, 9/4, 66/21 etc.

Hence the correct option is C.

In the following venn diagram N, S and M stands for Nokia, Sony Ericsson and Motorola respectively.

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Given that (a − c) : (a − b) = 2 : 3 and a : c = 2 : 1. Let us assume that a = 4K, b = K and c = 2K where K is a constant

Now, 4K + 50 − x + 2K + 25 − x + K = 153

Or

K=78+2x

The fallowing are the possible values of K and x that satisfy the above equation: (K = 12, x = 3) ; (K = 14, x = 10) and (K = 16 and x = 17)

Minimum possible number of persons who bought Nokia a + (30 - x) + x + ( 20 - x) = 50 + 4K - x = 50 + (4 x 16) - 17 = 97.

AO = BO(radius) and CA = BD(given)

Adding the above equation

AO + AC = BO + BD

OC = OD

∠ODC = ∠OCD

∠COD = 90^∘

Therefore

∠ODC = ∠OCD = 45^∘

Join OP.

OP is perpendicular to CD. (Line joining center to tangent at point of contact)

Tan ∠ODP = OP/PD

tan⁡45^∘ = 100/PD

1 = 100/PD

⇒ PD = 100cm

sin⁡ ∠ODP = OP/PD

sin⁡45^∘ = 100/OD

OD = 100/0.7071

⇒ OD = 141.42cm

BD = OD + OB

⇒ BD = 141.42 − 100

⇒ BD = 41.42cm

∴ AC + CP = 100 + 41.42 = 141.42cm

Hence the correct option is C.

We have

OA ⊥ AP and OB ⊥ BP [∴ the tangent at any point of a circle is perpendicular to the radius through the point of contact] Join OP.