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Algebra: Basics Test-1
Question 1 |
If a× b = 2a-3b+ab, then 3 × 5 +5 x 3 is equal to:
20 | |
23 | |
24 | |
22 |
Question 1 Explanation:
We are given by the statement that
a×b=2a−3b+ab⇒3×5=2×3−3×5+3×5=65×3=2×5−3×3+3×5=10−9+15=16∴3×5+5×3=6+16=22
a×b=2a−3b+ab⇒3×5=2×3−3×5+3×5=65×3=2×5−3×3+3×5=10−9+15=16∴3×5+5×3=6+16=22
Question 2 |
If p ×q = p × q + p/q , the value of 8 ×2 are:
13 | |
14 | |
15 | |
16 |
Question 2 Explanation:
p×q=p+q+pq∴8×2=8+2+82=10+4=14
Question 3 |
Two numbers a and b ( a > b) are such that their sum is equal to three times their difference.Then value of 3ab2(a2−b2) will be:
112 | |
123 | |
113 | |
1 |
Question 3 Explanation:
From the given statement we can say that
(a+b)=3(a−b)=3a−3b⇒3b+b=3a−ra⇒2a=4b⇒a=2b⇒ab=21⇒ab=12∴a=2,b=13ab2(a2−b2)=3×2×12×(4−1)=66=1
(a+b)=3(a−b)=3a−3b⇒3b+b=3a−ra⇒2a=4b⇒a=2b⇒ab=21⇒ab=12∴a=2,b=13ab2(a2−b2)=3×2×12×(4−1)=66=1
Question 4 |
The value of <br>
(1+1p)(11p+1)(1+1p+2)(1+1p+3)
is
is
1+1p+4 | |
p+4 | |
p+4p | |
1p |
Question 4 Explanation:
We can solve the given fraction as
=(1+1p)(1+1p+1)(1+1p+2)(1+1p+3)=p+1p×p+2p+1×p+3p+2×p+4p+3=p+4p
=(1+1p)(1+1p+1)(1+1p+2)(1+1p+3)=p+1p×p+2p+1×p+3p+2×p+4p+3=p+4p
Question 5 |
If a × b = 2 (a + b), then 5 ×2 is equal to:
16 | |
12 | |
14 | |
18 |
Question 5 Explanation:
Given statement is
a×b=2(a+b)∴5×2=2(5+2)=2×7=14
a×b=2(a+b)∴5×2=2(5+2)=2×7=14
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