Q1. The father’s age is 6 times his son’s age. After 4yr, the age of the father will be four times his son’s age. The present ages in years of the son and father are, respectively –
(a) 4 and 24
(b) 5 and 30
(c) 6 and 36
(d) 3 and 24
Sol.
Let son’s age = x yr.
Let son’s age = x yr.
Then, 6x + 4 = 4(x + 4)
⇒ 2x = 12
∴ x = 6 yr.
Q2. If (x – 2) if a factor of x^3+ax^2+bx+16 and b = 4a, then respective values of a and b are respectively –
(a) 2, –8
(b) –2, 8
(c) –2, –8
(d) 2, 8
Sol.
Let f (x) = x^3+ax^2+bx+16
Let f (x) = x^3+ax^2+bx+16
Since, (x – 2) is a factor of f(x).
∴ f (2) = 0
i.e. 8 + 4a + 2b + 16 = 0
= 8 + 4a + 8a + 16 = 0
a= –2
Q3. If the area of an equilateral triangle is 16√3 cm^2, then the perimeter of the triangle is –
(a) 48 cm
(b) 24 cm
(c) 12 cm
(d) 36 cm
Sol.
Let side of equilateral triangle = a cm
Let side of equilateral triangle = a cm
∴ √3/4 a^2=16 √3
⇒ a = 8 cm
∴ Perimeter = 3a = 24 cm
Q4. The polynomial
P(x)=x^4-2x^3+3x^2-ax+3a-7
When divided by (x + 1) leaves the remainder 19. The value of a is –
(a) 3
(b) 2
(c) 5
(d) 4
Sol.
P (x) – (x + 1) .g (x) + 19
P (x) – (x + 1) .g (x) + 19
⇒ P(x) – 19 = (x + 1). g(x)
i.e. (x + 1) divides p (x) – 19 completely.
i.e. P(–1) – 19 = 0
⇒ [1 – 2(–1) + 3 – a (–1) + 3a – 7] – 19 = 0
⇒ 4a = 20
∴ a = 5
Q5. Four traffic lights give red signals after 30 s, 45 s, 60 s and 120 s respectively. If all the traffic lights began at the same time, then all of them will give red signal at the same time in –
(a) 180 s
(b) 120 s
(c) 300 s
(d) 360 s
Sol.
required solution = LCM (30, 45, 60, 120) = 360 s
required solution = LCM (30, 45, 60, 120) = 360 s
Q6. O is the centre of circle in following figure. ∠OCB equals –
(a) 10°
(b) 30°
(c) 20°
(d) 40°
Sol.
Since opposite angles of a cyclic quadrilateral are supplementary,
Since opposite angles of a cyclic quadrilateral are supplementary,
∴ ∠BAC+∠BDC=180°
⇒ ∠BDC=180°-∠BCA=80°
Now, ∠BOC=2∠BDC=160°
In ∆OBC,OB=OC (radius of the circle)
∴ ∠OBC=∠OCB
∠OBC+∠OCB=180°-∠BOC
⇒ 2∠OCB=20°
⇒ ∠OCB=10°
Q7. Two cylinders have equal volumes and their heights are in the ratio 1 : 3. The ratio of their radii will be –
(a) 1: 3
(b) 3: 1
(c) √3 : 3
(d) 3: √3
Sol.
πr_1^2 h^1=πr_2^2 h^2
⇒ (r_1^2)/(r_2^2 )=h_2/h_1 =3/1
⇒ r_1/r_2 =√3=3/√3
Q8. Area of two similar triangles are 4 cm^2 and 9 cm^2 respectively. The ratio of their corresponding sides is –
(a) 4 : 9
(b) 9 : 4
(c) 3 : 2
(d) 2 : 3
Sol.
(Area (∆1 ))/(Area (∆2))=((Side (∆1 ))/(Corresponding side (∆2 ) ))^2
⇒ 4/9=(s1^2)/(s2^2 )
⇒ s1/s2 =2/3
Q9. A person have some one rupee, 50 paisa and 25 paisa coins totaling Rs. 210 in the ratio 5: 6: 8, then number of one rupee coins is –
(a) 41
(b) 42
(c) 103
(d) 105
Sol.
The ratio of the number of Rs. 1 : 50 paise and 25 paise coins is
The ratio of the number of Rs. 1 : 50 paise and 25 paise coins is
5 : 6 : 8
Therefore, the ratio of their contribution in Rs. 210 is
5∶6/2:8/4 i.e. 5∶3∶2
∴ Contribution of Rs. 1 coins in Rs. 210
=(5/(5 + 3 + 2))×210
= Rs. 105
∴ Number of Rs. 1 coins = 105
Q10. If 1/x+1/y=u and xy=1/v , then the value of 1/x^2 +1/y^2 is –
(a) 1/u^2 -2/v
(b) u^2-2/v
(c) u^2-2v
(d) 1/u^2 -2v
Sol.
(1/x+1/y)^2=u^2
(1/x+1/y)^2=u^2
⇒ 1/x^2 +1/y^2 +2/xy=u^2
⇒ 1/x^2 +1/y^2 =u^2-2/v
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