CUET2024-General Test-28/05/2024
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आज की पोस्ट इसप्रकार है:
Q 2605) Two numbers are in the ratio 9:11. If the sum of numbers is 200, then their difference will be —
(A) 15 (B) 18 (C) 20 (D) 23
Ans: (C) 20
Solution:
Suppose, the two numbers are 9y and 11y.
According to the question, 9y+11y=200; or, 20y=200; or, y=200/20 =10
Therefore, two numbers are, 9×10=90 and 11×10=110
Therefore, the difference of number is 110-90=20
Q 2606) The value of (9)3+(-17)3+(8)3 is —
(A) 3636 (B) 2672 (C) 3672 (D) -3672
Ans: (D) -3672
Solution:
We know that, a3 +b3 +c3 = 3abc if a+b+c=0
Here, a=9 + b=-17 + c=8 ; Total =0
Therefore, 3(9) (-17) (8) =-3672
Q 2607) The value of (1.7)3-(1.2)3/(1.7)2+1.7×1.2+(1.2)2
(A) 0.5 (B) 0.05 (C) 2.9 (D) 0.29
Ans: (A) 0.5
Solution:
We know that, a3-b3 =(a-b) (a2+ab+b2)
or, a3-b3/(a2+ab+b2) =(a-b)
Where, a=1.7 and b=1.2
Therefore, (1.7-1.2)(1.7)3-(1.2)3/(1.7)2+1.7×1.2+(1.2)2 = (1.7-1.2)
or, (1.7)3-(1.2)3/(1.7)2+1.7×1.2+(1.2)2 =0.5
Q 2608) The value of (1.7)3+(1.2)3/(1.7)2-1.7×1.2+(1.2)2
(A) 0.5 (B) 29 (C) 0.29 (D) 2.9
Ans: (D) 2.9
Solution:
We know that, a3+b3 =(a+b) (a2-ab+b2)
or, a3+b3/(a2-ab+b2) =(a+b)
Where, a=1.7 and b=1.2
Therefore, 1.73+1.23/(1.7)2-1.7×1.2+(1.2)2=(1.7+1.2)
or, 1.73+1.23/(1.7)2-1.7×1.2+(1.2)2 =2.9
Q 2609) If 6 men can do a piece of work in 10 days, then in how many days 4 men will do the same work ?
(A) 7 days (B) 8 days (C) 10 days (D) 12 days
Ans: (C) 10 days
Solution:
6 men can do a piece of work in 10 days
Therefore, one man can do the same work = 6×10=60 days.
Therefore, 4 men can do the same work = 60/4 =15 days.
Q 2610) In the given figure, PQ ⊥ PR, PQ II RS, ∠RQS=40° and ∠QST=75°, then z is —

(A) 55° (B) 60° (C) 65° (D) 68°
Ans: (A) 55°
Solution:
From the given figure PQ II RS
In ∆QRS, y+105°+40°=180°; or, y=180°-145°=35°
Therefore, In ∆PRQ, z+35°+90°=180°; or, z=180°-125°=55°
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General Test
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Vimal Kumar Tulsyan is the Founder of CUET NOW, an educational platform focused on CUET UG preparation. He has more than 10 years of teaching experience in Reasoning and General Aptitude.
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