**2019-01-16 13:00:41.0**

**Ex: The average value of property of Sweta, Anushika and Anjali is Rs.130 lakhs. The property value of Sweta is 20 lakhs greater than the property value of Anushika and Anjali's property value is 50 lakhs greater than the Sweta's property value. The value of property of Anjali is-**

(A) Rs.120 lakhs

(B) Rs.170 lakhs

(C) Rs.150 lakhs

(D) Rs.100 lakhs

**Correct answer: B**

**Trick:**

Property value of Anushika is Rs.x

130 × 3 = x + x + 20 + x + 20 + 50

390 = 3x + 90

3x = 300

x = 100

Anjali = 100 + 20 + 50 = 170 lakhs

**2019-01-15 12:16:35.0**

**Ex: The compound interest on Rs.8000 at 15% per annum for 2 years 4 months, compounded annually is -**

(A) Rs.2980

(B) Rs.3091

(C) Rs.3109

(D) Rs.3100

**Correct answer: C**

**Trick:**

Required Compound interest = 1200 + 1200 + 180 + 529

= Rs.3109

**2019-01-14 12:51:49.0**

**Ex: A sum of money at simple interest amount to Rs.8118 in 3 years and to Rs.9449 in 4 years. The sum is-**

(A) Rs.4135

(B) Rs.4130

(C) Rs.4120

(D) Rs.4125

**Correct answer: D**

**Trick:**

Simple interest = 9449 – 8118 = Rs.1331

Simple interest in 3 years = 3 × 1331

= Rs.3993

Sum = 8118 – 3993 = Rs.4125

**2019-01-09 12:57:22.0**

**Ex: When a sphere is cut into two hemisphere the total surface area of two hemisphere is equal to the area of circle whose radius is equal to the side of square whose area is 144 cm**

**
****Ex: A sum of money at simple interest amount to Rs.8118 in 3 years and to Rs.9449 in 4 years. The sum is-**

**2019-01-09 11:33:50.0**

**Ex: When a sphere is cut into two hemisphere the total surface area of two hemisphere is equal to the area of circle whose radius is equal to the side of square whose area is 144 cm². Then find the radius of sphere ?**

A. 2√3 cm

B. 3√6 cm

C. 8 cm

D. 2√6 cm

**Correct answer: D**

**Trick:**

Let radius of circle be r cm.

∴ r = a = √144 = 12 cm

ATQ,

Let radius of hemisphere be R.

3πR² + 3πR² = πr²

6πR² = π × 12 × 12

R = √24 = 2√6 cm

**2019-01-08 12:38:17.0**

(A) Rs.4135

(B) Rs.4130

(C) Rs.4120

(D) Rs.4125

**Correct answer: B**

**Trick:**

Simple interest = 9449 – 8118 = Rs.1331

Simple interest in 3 years = 3 × 1331

= Rs.3993

Sum = 8118 – 3993 = Rs.4125

**2019-01-07 11:56:31.0**

**Ex: P is thrice more as good as a workman Q. They together can do a job in 16 days. In how many days will Q finish the work?**

(A) 32 days

(B) 40 days

(C) 60 days

(D) 64 days

**Correct answer: D**

**Trick:**

P is thrice as good as Q

(P + Q) = 3Q + Q = 4Q

4Q complete job in 16 days

Q will complete the job in = 4 × 16 = 64 days

**2019-01-04 17:07:03.0**

**Ex: If φ is a positive acute angle and 4cos**

^{2}φ – 4cosφ + 1 = 0, then the value of tan(φ – 15^{o}) is equal to-
(A) 0

(B) 1

(C) √3

(D) 1/√3

**Correct answer: B**

**Trick:**

4 cos

^{2}φ – 4 cos φ + 1 = 0
(2cos φ – 1)

^{2}= 0
2cos φ – 1 = 0

cos φ = = cos60o

φ = 60

^{o}
tan (60

^{o}– 15^{o}) = tan45^{o}= 1**2019-01-03 16:02:23.0**

**Ex: If the profit percent got on selling an article is numerically equal to its cost price in rupees and the selling price is Rs. 39, then cost price (in Rs.) will be-**

(A) 20

(B) 22

(C) 28

(D) 30

**Correct answer: D**

**Trick:**

Let the C.P. of the article be Rs.x/माना वस्तु का क्रय मूल्य x रु. है ।

Gain/लाभ % = x%

39 - x / x * 100 = x

x

^{2}= 3900 – 100x
x

^{2}+ 100x – 3900 = 0
x

^{2}+ 130x – 30x – 3900 = 0
x(x + 130) – 30 (x + 130) = 0

x= 30, – 130

**2019-01-02 11:54:44.0**

**Ex: A field is 100 metre long and 60 metre wide. A path of uniform width runs round it on the inside, if the area of the path is 1500 m2. Find the width of the path.**

(A) 5

(B) 4

(C) 2.5

(D) 3.5

**Correct answer: A**

**Trick:**

Let width of path = x m

100 × 60 – (100 – 2x) (60 – 2x) = 1500

– 4x2 + 320 x – 1500 = 0

x2 – 80x + 375 = 0

x = 5, 75

x = 5