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The greatest number which when divide 80,121 & 148 leave remainder 2, 4 & 5 respectively = HCF (80 - 2, 121 -4 , 148 - 5) = HCF (78, 117, 143) = 13
25% of 280 × 15% of 460 = ? × 230
If 1210 ÷ 22 + 1332 ÷ 37 - y + 54 × 3 = 980 ÷ 20 × 144 ÷ 48, then the value of y is:
(7/5) × (3/4) × (5/9) × (6/7) × 3112 = ?
{(? × 15) + (? × 45)} – 120 = 360
√676 + (0.75 × 80) + (72 ÷ 3) = ? - 82
(21 × 51 + 81)/(9 × 14 - 30) = ?
25.6% of 250 + √? = 119
120% of 400 + ?% of 520 = 1000
140% of 1270 + 60% of 2085 = 1881 + ‘?’% of 287
