## Advance Maths Questions for SSC CGL CAT

Q1. Consider an obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?

A. 5
B. 21
C. 10
D. 15

Q.2. In the given figure , PQR is a triangle and quadrilateral ABCD is inscribed in it . QD = 2 cm , QC = 5 cm , CR = 3 cm , BR = 4 cm , PB = 6 cm , PA = 5 cm and AD = 3 cm . What is the area ( in cm² ) of the quadrilateral ABCD?

A. 17√21
B. 17/5
C. 17√21/5
D. √21/5

Q3. The quadratic equation x2 + mx + n = 0 has roots that are twice those of x2 + px + m = 0, and none of m, n and p is zero. What is the value of n/p?
(A) 2
(B) 16
(C) 4
(D) 8

Q4. Akash when going slower by 15 Km/hr, reaches late by 45 hours. If he goes faster by 10 Km/hr from his original speed, he reaches early by by 20 hours than the original time. Find the distance he covers.
आकाश जब 15 किमी/घंटा तक धीमी गति से चलता है, तो 45 घंटे देरी से पहुंचता है. यदि वह अपनी मूल गति से 10 किमी/घंटा तेजी से चलता है, तो वह मूल समय की तुलना में 20 घंटे जल्दी पहुंचता है. उसके द्वारा तय की जाने वाली दूरी ज्ञात कीजिये?
(A) 8750 Km/किमी
(B) 9750 Km/किमी
(C) 1000 Km/किमी
(D) 3750 Km/किमी

Solution

1. Solution
If c is the side opposite the obtuse angle & hence the longest, it should satisfy the following two conditions:

a2 + b2 < c2 and c < a + b

If 15 is the longest side, then:

82 + x2 < 152
64 + x2 < 225
x2 < 161
x < 12.69

So, keeping with our other range, x could be 8, 9, 10, 11 or 12.

If we make x our longest side, we have:

82 + 152 < x2
64 + 225 < x2
289 < x2
17 < x

x could be 18, 19, 20, 21 or 22.

So 10 possible values for x.

Ans. C

2. Solution
Area of the ΔPQR = √(14×6×4×4) = 8√21

By using sin formula for area = 1/2abSinθ

Ratio of area of ΔAPB and ΔPQR = 5×6/10×10 = 30/100 = 3/10 = 3×8/10×8 = 24/80

Ratio of area of ΔQDC and ΔPQR = 5×2/10×8 = 10/80

Ratio of area of ΔBRC and ΔPQR = 3×4/10×8 = 12/80

So the ratio of area of □ABCD and ΔPQR = (80 - 24 - 10 - 12)/80 = 34/80

Now the area of 80 unit ΔPQR = 8√21

So 1 unit = 8√21/80 = √21/10

So the area of 34 unit □ABCD = 34×√21/10

= 17√21/5 Ans.

3. Solution

4. Solution
Let the actual speed is "v" distance "d"
According to the conditions:
d/(v - 15) = d/v + 45
=> 15d/v(v -15) = 45  .....(i)
d/(v + 10) = d/v - 20
=> 10d/v(v +10) = 20  .....(ii)
divide (i) by (ii)
15(v + 10)/10(v - 15) = 45/20
(v + 10)/(v - 15) = 3/2
v = 65
from equation (i)
15d/65×50 = 45
d = 9750
So the total distance will be 9750 km

Advance Maths Questions for SSC CGL CAT Reviewed by Super Pathshala on 13:18 Rating: 5