# WBJEE 2015 Solved Mathematics Question Paper – Part 7

Find **WBJEE 2015 Solved Mathematics Question Paper – Part 7** in this article. This paper consists of 10 questions (#61 to #70) from WBJEE 2015 Mathematics paper. Detailed solution of these questions has been provided so that students can match their solutions.

**Importance of Previous Years’ Paper:**

Previous years’ question papers help aspirants in understanding exam pattern, question format, important topics and assessing preparation. It has also been seen that sometimes questions are repeated in WBJEE Exam. So, this paper will certainly boost your confidence.

**About WBJEE Exam**

WBJEE is a common entrance examinations held at state level for admission to the Undergraduate Level Engineering and Medical courses in the State of West Bengal. The WBJEE engineering entrance exam has two sections – Mathematics, Physics and Chemistry. The Mathematics section of WBJEE 2015 engineering entrance exam consists of 80 questions.

(A) b + c

(B) c + a

(C) a + b

(D) a + b + c

**Correct Option: (C)**

**Sol.**

In the above figure, OECD is a square since all angles are right angle.

So,

BE = a – r

And, AD = b – r

Also, tangents drawn from an external point are equal.

So,

BF = BE

And, AF = AD

(A) a + b = c

(B) b + c = a

(C) c + a = b

(D) c = a

**Correct Option: (D)**

**Sol.**

We know that,

Now putting above values in the given equation,

Then sum of the elements of *U ^{–1} *is

(A) 6

(B) 0

(C) 1

(D) 2/3

**Correct Option: (B)**

**Sol.**

**WBJEE 2016 Solved Physics and Chemistry Question Paper**

(A) 1000

(B) 500

(C) 1/500

(D) 1/1000

**Correct Option: (D)**

**Sol.**

**65.** If 5 distinct balls are placed at random into 5 cells, then the probability that exactly one cell remains empty is

(A) 48/125

(B) 12/125

(C) 8/125

(D) 1/125

**Correct Option: (A)**

**Sol. **

If we have to keep exactly one cell empty, then one of the cell must contain 2 balls and the remaining cell must contain one ball each.

Now let us assume two balls as one unit.

So, we have now 4 units to be arranged in 4 places.

This can be done in 4! ways.

**66.** A survey of people in a given region showed that 20% were smokers. The probability of death due to lung cancer, given that a person smoked, was 10 times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is 0.006, what is the probability of death due to lung cancer given that a person is a smoker?

(A) 1/140

(B) 1/70

(C) 3/140

(D) 1/10

**Correct Option: (C)**

**Sol.**

Let us assume that,

S = person is smoker

NS = person is non smoker

D = death due to lung cancer

**67.** A person goes to office by a car or scooter or bus or train, probability of which are 1/7, 3/7, 2/7 and 1/7 respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is 2/9, 1/9, 4/9, and 1/9 respectively.

Given that he reached office in time, the probability that he travelled by a car is

(A) 1/7

(B) 2/7

(C) 3/7

(D) 4/7

**Correct Option : (A)**

**Sol. **

## WBJEE Sample Question Paper Set-II

(A) 2015

(B) 0

(C) 2015 × 2016

(D) 2015 × 2014

**Correct Option: (C)**

**Sol. **

** **Applying L’ Hospital Rule,

**70.** If x and y are digits such that 17! = 3556xy428096000, then x + y equals

(A) 15

(B) 6

(C) 12

(D) 13

**Correct Option : (A)**

**Sol. **

Since 17! is divisible by all numbers less than or equal to 17.

So, 17! is also be divisible by 9.

So,

3+5+5+6+x+y+4+2+8+0+9+6+0+0+0 is divisible by 9

48 + x + y must be divisible by 9.

So, x + y can be 15 or 6.